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Internat. J. Math. & Math. Sci.
Vol. 8 No.4 (1985) 625-633
625
CONVEX CURVES OF BOUNDED TYPE
A. W. GOODMAN
Mathematics DepartmentUniversity of South Florida
Tampa, Florida 33620
(Received November 30, 1984)
ABSTRACT. Let C be a simple closed convex curve in the plane for which the radius of
curvature 0 is a continuous function of the arc length. Such a curve is called a convex
curve of bounded type, if 0 lies between two fixed positive bounds. Here we give a new
and simpler proof of Blaschke’s Rolling Theorem. We prove one new theorem and suggest
a number of open problems.
,," ;/{)!0:: ,I{7) i’IIRA,,<,’,V. Cooex cu,ve, boTxn.-icd type perimeter centnofd, BZaschkcs
/,’ 7in..{ eorem, pa,a el cue, ’as distri,ution on a curve.
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. INTRODUCTION.
Let C be a simple closed convex curve in the plane. Such curves have been the
subject of numerous studies[l, 2, 3, 5, 6, 7, 8, 12, 14, 15, 19, and 24] to cite only a
few. Here we will refine the objects of study by looking at certain subsets. Through-
out this paper C is a simple closed convex curve in the plane for which the radius of
curvature 0 is a continuous function of arc length. Our refinement consists of putting
upper and lower bounds on O.
Definition i. We say that C is a convex curve of bounded type if there are con-
stants R and R2 such that
0 R 0 R2 (I.I)
at every point of C. We let CV(RI,R2) denote the set of all such curves that satisfy
(1.1) for fixed R and R2.Theorems about the class CV(RI,R2) appear in the literature (see for example
Theorem 3), but as far as I am aware, this class has not been given a specific name and
symbol until now. In this work we are concerned with one type of question, namely how
close can C come to its "center" and how far away from its "center" can C go.
The center can be defined in various ways. For example the center of mass of the
region bounded by C when the region has a uniform mass distribution. Or the center
could be the center of mass of the curve C when the mass is distributed either uniformly
or as some other function of s the arc length on C. In any case we can take the origin
as the center of mass without loss of generality. For each fixed curve in CV(RI,R2) set
D min IOPI, and D2 maxlOPl (1.2)PC PC
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626 A.W. GOODMAN
Our main result is
Theorem I. Suppose that C CV(R I, R2) and the center is the center of mass of
the curve C. If the mass distribution on C is uniform, then
R =< I) _<_ D 2 =< R2 (1.3)
The two circles of radius R and R2 show that the inequality (1.3) is sharp.
Bose and Roy [6] call this center the perimeter centroid.
in section we review some facts about parallel curves and we give a new proof of
Blaschke’s Roiling Theorem [3, pp. 114-116]. In section 3 we prove Theorem 1. In
section 4 we suggest some topics for further research on the set GV(R1,R2).2. PARALLEL CURVES.
Let C e CV(R1,R2). We select the parameter s (arc length) so that s increases as
the point P P(s) traverses C in the counterclockwise direction. Let 0 denote, as
usual, the angle that the unit tangent T makes with the positive x-axis, and let N be
the unit inward normal to C at the point P P(s). Ue recall that
dx dycos 0, sin 0 (2 I)
ds ds
dxi dd-s _,Tds-
+ J- (cos 0)_i + (sin 0)j (2.2)
and
N (-sin 0)i + (cos 0)j. (2.3)
If V V(s) is the vector equation of C we introduce a second curve C* defined by
and
dy*= dy A< sin 0 (l-A<)sin 0. (2 6)ds ds
We let s*, *, and 0* denote, arc length, curvature, and radius of curvature at the
corresponding point on C*. Then (2.5) and (2.6) give
ds ] \ds ]+ (I-A) 2. (2.7)
if R A R2, then the curve C* may have cusps as shown in Fig. I. If A R
we set ds*/ds A 0. If A R2 then A 0 and we set ds*/ds
Thus in either case s* and s increase together. In the first case, A < R I, we havedV* d* dsT* [(I-A)cos 0 i + (l-A)sin 0 j] I-Adq* ds ds*
(cos )i + (sin )j T. (2.8)
If A R2, then the same type computation gives * -.Lemma I. If A R I, then the directed tangents at corresponding points of C and
C* are parallel and point in the same direction. Further N_* . If A R2, then T_*
-T and N* -N.
the vector equation V* V(s) + AN, where A is a constant. The curve C* is said to be
parallel to C, see [13 pp. 80-84, 18 p. 67, and 19]. Fig. shows a number of curves
parallel to the ellipse 2/9 + y2/4 I. The curve C* is also a Bertrand mate of C,
although the term Bertrand curve usually refers to twisted curves in space [4, p. 35].
If P(x,y) is a point on C and P*(x*,y*) is the corresponding point on the parallel
curve C*, then
x* x A sin and y* y + A cos 0. (2.4)
If I/0 is the curvature of C at P, then < d0/ds and from (2.4) and (2.1)
dx* dxds ds
A< cos (1-A)cos , (2.5)
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CONVEX CURVES OF BOUNDED TYPE 627
A=O
=2
\
"’ + ’’,,
Figure
Lemma 2. If C CV(RI, R2) and A < RI, then Ca is locally convex and at corres-
ponding points 0 n 0 A.
By locally convex we mean *’ (s*) 0 at each point of Ca.Proof. From Lemma we have * at corresponding points. Hence, for the
curvature
, d* d d dsds* ds* ds ds* I-A/o"
Thus * 0 whenever O, and C* is locally convex. Further
(2.9)
0* I-A/00 (i A) 0 A (2 I0),
Of course 0 n 0 A is geometrically obvious from the definition of Ca. Q.E.D.
If A R2, the factor 1/(I-A/o) in (2.9) is replaced by 0/(A-0). Again Ca is
locally convex, but in this case 0* A 0.
It is geometrically obvious that if A R or A R2, then C* is a simple closed
curve. It seems that a direct proof is rather elusive. The difficulty may lie in the
following example. Let Ca be the image of Izl under the complex function f(z)
z + z 2. Then Ca is convex in the sense that * 0 at every point, so C* is locally
convex. But this curve fails to be simple. Nevertheless we have
Theorem 2. If C E CV(RI,R2) and A R or A R2, then C* is a simple closed
convex curve. If A RI, then Ca CV(RI*, R2*), where
RI* R A, and R2* R2 A. (2.11)
If A R2, then Ca g CV(RI* R2*), where
RI* A R2, and R2* A RI. (2.12)
Proof. We have already seen that C’is locally convex, but the example shows this
is not sufficient to prove that Ca is simple. On Ca letL*
A* *(e*) *(0) f ds*, (2.13)0 os
where L* is the length of Ca. We make a change of variables from s* to s. If A < R I,
then * and
d* dds ds"
Then (2.13) gives
L* L L d@ L, f d* ds ds* f d* ds f ds f d 27.0 d- d-* 0 d- 0 s 0
(2.14)
(2.15)
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628 A.W.GOODMAN
Since C* is locally convex and A* 2, we see that C* is a simple curve.
If A R2, then * + . Hence (2.14) is still true and the proof remains valid.
The relations (2.11) and (2.12) follow from 0* 0 A and O* A 0 respectively. Q.E.D.
Theorem 3. Let C CV(R1,R2) and let K be a circle tangent internally to C at any
point PO of C. If K has radius RI, then K is contained in C. If K has radius R2, then
K contains C.
This theorem is often called Blaschke’s Rolling Theorem, because it states that
(a) a circle of radius R can roll around the inside of C, and (b) a circle of radius
R2 can roll around the outside of C. Blaschke has extended his theorem to 3-dimensional
space [3, p. 118]. For further work on this theorem, and various extensions see [II, 17,
20, and 22].
To be precise the phrase "internally tangent" means that K is tangent to C at POand the center of K lies on the inward normal to C at P0" Thus the location of the
center is given by equation set (2.4) with A replaced by R= the radius of the tangent
circle (a 1,2). We say that K is contained in C if K is contained in the closure of
the region bounded by C. Further K contains C, if C is in the closed disk bounded by K.
Proof of Theorem 3. We first show that the curve C cannot cross the circle K in a
neighborhood of P0’ the point of contact. Without loss of generality let PO be the
origin and let K and C be tangent to the x-axis at the origin. Further suppose that
both the circle and the curve lie above the x-axis, except at the origin. In this
position the lower half of the circle will have equation
Y R /R2_x2 -R x _< R. (2.16)
If y f(x) is the equation of C in a suitable neighborhood, I -e < x < e, then we
have f’(x) sgn x 0 and f’’(x) > 0 in I.
Lemma 3. Suppose that 0 R in I, where 0 is the radius of curvature on C. Then,
under the conditions described above
y(x) Y(x) R /R2 x2 x g I.
Thus in I, the curve C cannot cross from outside to inside K, but of course C may
coincide with K. We omit the proof of Lemma 3, but it follows directly from two inte-
grations, starting with the inequality
y’’(x)< (2.17)
3/2 R[l+(y’(x)) 2]
By reversing the inequality signs we have
Lemma 4. Suppose that 0 R in I. Then under the conditions on K and C des-
cribed above
y(x) J Y(x) R- /_X x e 1
Thus in I, the curve C cannot cross from inside to outside K, but of course C may
coincide with K.
From these two lemmas we see that if R R or R R2, then C cannot cross into or
out of K in a neighborhood of a point of tangency. To complete the proof of Theorem 3,
we must obtain this same result in the large.
First suppose that K has radius R and is tangent internally to C at PO" If K is
not contained in C, then K crosses C at a point P2 distinct from PO" Then we may find
a smaller circle K0 with radius R0 < R I, and such that K0 is tangent internally to C
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CONVEX CURVES OF BOUNDED TYPE 629
at PO’ and is tangent to C at another point PI’ see Fig. 2.
Po
Figure 2
C*Now consider the parallel curve with A R0 < RI. By Theorem 2, this curve is
a simple close curve. On the other hand, the center D of the circle K0 is at least a
double point of C* because it is the corresponding point for both P0 and PI" Hence we
have a contradiction.
For the second part of Theorem 3 let K be a circle with radius R2 and tangent in-
ternally to C at P0" If K does not contain C, then K crosses C at a point P2 distinct
from PO" Then we may find a larger circle K0 with radius R0 > R2 and such that K0 is
tangent internally to C at P0 and is tangent to C at another point PI" Again consider
the parallel curve C* with A R0 > R2. By Theorem 2 this curve C* is a simple closed
curve. Just as before we obtain a contradiction because D the center of K0 is at least
a double point on C*. Q.E.D.
Corollary I. Let L(C) denote the length of C and let A(C) denote the area of the
region enclosed by C. If C E CV(RI,R2), then
2RI ! L(C) ! 2R2, (2.18)
and
RI2 ! A(C) ! R22" (2.19)
The circles of radius R and R2 show that both of these inequalities are sharp.
The inequalities (2.18) and (2.19) are well known, see for example [I, p. 352],
[15], and [16, Vol. I, pp. 529 and 548].
3. PROOF OF THEOREM I.
Let C g CV(RI,R2) and let (s) be a mass distribution of C. We exclude the trivial
case in which all of the mass is concentrated at one point. Then the center of mass will
be an interior point of the region bounded by C. Without loss of generality we select
the center of mass to be the origin. If L is the length of C, then
IL Lx(s)(s)ds 0 and y(s)(s)ds 0 (3.1)
0 0
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630 A.W. GOODMAN
Now consider the parallel curve C* where A RI, and let * *(s*) be a mass distrl-
bution on C*. Then the moments Mx* and * are given by
* /L* x* (s*) * (s*) ds*0
and.L*
MX* 0 y*(s*)*(s*)ds*.
The change of variable from s* to s yelds
(3.2)
(3.3)
AIL (x- * (s*(s))(l-0-)ds, (3.4)M y* 0
andA
Mx* It (y+As) U* (s*(s))(l p--)ds. (3.5)
0
We now specialize, by setting (s) on C and selecting B*(s*) so that
u*(s*(s)) l-A/o(s) 0. (3.6)
Then (3.4) and (3.5) give
* IL (x-As)ds I
Lx ds A I dy 0,
0 0 C
and
Mx* (y+ ds y ds + A dx 00 C
from (3.1). Thus with the mass distribution (3.6), the center of mass of C* is also at
the origin. Since C* is a simple closed convex curve, the origin lies inside C* and
hence D A. Finally we note that A may be taken arbitrarily close to R E minlp for
points on C. Therefore
D RI. (3.7)
To prove that D2 R2, we consider the parallel curve C* where now A > R2. For
this curve equations (3.2)and (3.3) still hold. However, in this case we have
ds* A> 0. (3.8)ds
Thus in equations (3.4) and (3.5) we must replace the factor I- A/O by A/O I. If we
select (s) on C and u* on C A so that
(s*(s) o(s)A-o(s)
> O, (3.9)
then this mass distribution will give Mx* * 0. By Theorem 2, the curve C* is a
simple closed convex curve and the origin which is also the center of mass lles inside
the region bounded by C*.
No let P be a point on C furthest from the origin. Then OP is normal to C at P.If P* is the point on C* corresponding to P, then PP* is also normal to C at P. Hencethe points P, O, and P* are collinear.
Finally we observe that by Lemma I, the directed tangents to C and C* at the points
P and P* have opposite directions. Hence the origin is an interior point of the line
segment PP*. Therefore, IOP < IPP*I A. Since A may be taken arbitrarily close to
R2, we have D2 R2. Q.E.D.
4. FURTHER QUESTION FOR STUDY.
We observe that the inequality R D D2
R2
is sharp for the circles of
radius R and R2. But in each extreme case does not vary throughout the interval
[RI,R2] but instead is a constant at one end of the interval. The question naturally
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CONVEX CURVES OF BOUNDED TYPE 631
arises, can we find better bounds for D and D2
p is a continuous function of s
whose values fill out the interval [RI,R2]. A first candidate for consideration is
the ellipse x a cos t, y b sin t, 0 _< t _< 27, with 0 < b < a. If we set
b (RI2R2)3, and a (RIR22) 3, (4.1)
then p fills out the interval [RI,R2]. Further D b and D2 a, so the expressions
in (4.1) may appear as the proper lower and upper bounds for D. If true, this would
improve the bounds R and R2 given in Theorem I. However, by piecing together arcs of
circles, we can show that no better bounds than R < D < D2 < R2 can be obtained. To
see this, we give C only in the first quadrant and complete the curve by reflecting C
in the x- and yaxes.
Let C and C2 be the two arcs defined by
x a + R cos t, x R2 cos t,
y RlSin t, y -b + R2 sin t, (4.2)
0 < t < T, T < t 7/2,
respectively. The endpoints of the two arcs will meet when t T if we select a
(R2-RI) cos T and b (R2-R I) sin T where 0 < R < R2. If we compute the first deriva-
tive for the two arcs at t T, they will not be equal, but the tangent vectors will be
parallel, so that for this choice of a and b, the curve C C u C2 is a smooth curve.
Further p R on C and 0 R2 on C2. Finally D R sin T + R2(l-sin T)and DI/R as
T/ 7/2. Similarly D2 R2 cos T + R (l-cos T) and D2R2 as T/0. Thus no better bounds
than D2 R2 and D R can be proved under the hypotheses stated. Of course p is no___t
continuous in a neighborhood of P(T), where C and C2 meet, but it is merely a matter of
labor to alter the curve slightly at P(T) to make 0 continuous.
Perhaps some better bounds for D and D2 can be obtained if we impose a further re-
striction that the average values of 0 over the curve be a fixed number such as (RI+R2)/2.One can also examine the problem of finding sharp bounds for D and D2 i the mass
distribution has some fixed pattern, other than uniform. For example, Steiner [23],and [24, pp. 99-159] has considered curves in which the mass distribution on C is pr6-
portional to the curvature at each point of C. More generally one can select the mass
distribution to be some other function of I/0.
One can also consider Theorem i, when the center of mass of C is replaced by the
center of mass of the region enclosed by C. With this replacement, Theorem was proved
earlier by Nikllborc [21] and Blaschke [2]. It is reasonably clear that the center of
mass of a curve C is in general different from the center of mass of the region en-
closed by C, but it may be of interest to examine a particular example.
Let C be f(Izl=l) under f(z) z + az 2, where 0 < a < I/4. Then C is symmetric
with respect to the x-axis and if the mass distribution is uniform on C then the center
of mass will be on the x-axis. Hence it suffices to compute the x-coordinate. Let
d and C denote this coordinate for the domain center of mass and the curve center of
mass respectively. As easy computation gives
d a(4 3)
i+2a2"
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632 A.W. GOODMAN
A somewhat longer computation gives
MyRc --f-,
where
and
Hence
(4.4)
L f2/0 l+4a cos 0+4a2 dO, (4.5)
My /2(cos 0+a cos 20)/i+4a cos 0+4a2 dO (4 6)0
5 2+RC a(l- aIt is clear that in general d # C"
We may distinguish a third center of mass s, which we will call the conformal
strip center. Suppose that f(z) maps E conformally onto D, with f(0) O. Set Rs(r,l)the x-coordinate of the center of mass of the strip bounded by the curves
f(Izl=l) and f(IzI=r), where r I. Then by definition
s lim s(r,l). (4.7)r+l-
An easy computation shows that if f(z) z + az 2, 0 < a < I/4, and the mass distribu-
tion is uniform, then
s 2a
l+4a2 (4.8)
In this case s # d unless a 0. Further it is clear that in general s # C" This
example suggests the problem of finding
max lj-kl (4.9)
when C varies over the set CV(RI2) and j,k e {d,C,s}.
For other relations among various centers of mass, see Guggenheimer [I0], and Kubota
[19].
A computation, using Izl and
l’zf’(z)l (4 I0)P Re(l+zf"’(z)/’(z))’shows that for 0 < a < I/4, the function f(z) z + az 2 gives a convex curve
for which the radius of curvature is3/2
(I+4a cos B+4a2)l+6a cos 8+8a2 (4.11)
Extreme values of p occur when B 0, B , and cos B 2a. Thus z + az 2 is in
CV(RI,R2) for
R1 /l-4a2, and R2 (l-2a)2/(l-4a).One can also investigate the properties of normalized univalent functions that map
the unit disk conformally onto a region bounded by a curve in CV(RI,R2). Some elemen-
tary results in this direction have been obtained by the author [9].
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CONVEX CURVES OF BOUNDED TYPE 633
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